Answer:
The acceleration vector always points towards the centre of the motion. This is referred to as centripetal acceleration. It accounts for the centripetal force that tends to pull the object in circular motion towards the centre of motion.
Consider the diagram below;
it illustrates a body in circular motion along the black circle, with the acceleration vector a shown by the orange arrow pointing towards the centre of motion and the velocity vector v shown by the green arrow pointing tangentially away from the circumference.
Answer:
Tension of 132N
Explanation:
We need to apply Summatory of Force to find the tension in the hand.
We define te tensión in the hand as and the Tension in fence post as
, then
We apply summatory of moments then
Where the Force 2 is 1.25m from the center of summatory,
We can note that,
We have two equation and two incognites, then replacing (1) in (2)
Answer:
A) = 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = + m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase